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marysya [2.9K]
2 years ago
10

3 waves are shown with a line through their center. The bottom of the first wave is labeled C. A bracket labeled D connects the

bottom of the second and third waves. A line from the center to the top of the first wave is labeled A. The line leading to the top is labeled B.
Label the parts of this wave.
Chemistry
2 answers:
Likurg_2 [28]2 years ago
8 0

Answer:Label the parts of this wave.

A:  

✔ crest

B:  

✔ amplitude

C:  

✔ trough

D:  

✔ wavelength

Explanation:

emmasim [6.3K]2 years ago
8 0

A:  crest

B:  amplitude

C:   trough

D:   wavelength

i did the assignment and got it right.

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What is the empirical formula of a compound that is composed of 60.94% carbon 15.36% hydrogen and 23.70% nitrogen
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Answer:

C₃H₉N

Explanation:

The empirical formula of a compound is the fundamental and basic possible formula that shows the mole ratio of the atoms of each element in a molecule of the compound.

mole ratio of carbon = 60.94/12 = 5.078

mole ratio of hydrogen = 15.36/1  = 15.36

mole ratio of nitrogen = 23.70/14 = 1.693

Now; we will divide by the smallest value

So; carbon = 5.078/1.693 = 2.99 ≅ 3.0

hydrogen = 15.36/1.693 = 9.07 ≅ 9.0

nitrogen = 1.693/1.693 = 1 ≅ 1

Thus,  the empirical formula is = C₃H₉N

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2 years ago
Given the partial equation:
Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

First balance the main element in the reaction.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

Now balance oxygen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-+3H_2O

Now balance hydrogen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-+6H^+\rightarrow I^-+3H_2O

Now balance the charge.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}+2e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : 2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

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